# Simplification of Algebraic Fractions

Here we will learn simplification of algebraic fractions to its lowest term.

1. Simplify the algebraic fraction:

$$\frac{8a^{2}b}{4a^{2} + 6ab}$$

Solution:

$$\frac{8a^{2}b}{4a^{2} + 6ab}$$

We see in the given fraction the numerator is monomial and the denominator is binomial, which can be factorized.

$$\frac{\not{2}\times 2\times 2\times \not{a}\times a\times b}{\not{2}\not{a}(2a + 3b)}$$

We can see that ‘2’ and ‘a’ are the common factors in the numerator and denominator so, we cancel the common factor ‘2’ and ‘a' from the numerator and denominator.

= $$\frac{4ab}{(2a + 3b)}$$

2. Reduce the algebraic fraction to its lowest term:

$$\frac{x^{2} + 8x + 12}{x^{2} - 4}$$

Solution:

$$\frac{x^{2} + 8x + 12}{x^{2} - 4}$$

Each of the numerator and denominator is polynomial, which can be factorized.

= $$\frac{x^{2} + 6x + 2x + 12}{(x)^{2} - (2)^{2}}$$

= $$\frac{x(x + 6 ) + 2(x + 6)}{(x + 2)(x - 2)}$$

= $$\frac{(x + 2)(x + 6)}{(x + 2)(x - 2)}$$

We observed that in the numerator and denominator (x + 2) is the common factor and there is no other common factor. Now, we cancel the common factor from the numerator and denominator.

= $$\frac{(x + 6)}{(x - 2)}$$

3. Reduce the algebraic fraction to its lowest form:

$$\frac{5x^{2} - 45}{x^{2} - x - 12}$$

Solution:

$$\frac{5x^{2} - 45}{x^{2} - x - 12}$$

Each of the numerator and denominator is polynomial, which can be factorized.

= $$\frac{5(x^{2} - 9)}{x^{2} - 4x + 3x - 12}$$

= $$\frac{5[(x)^{2} - (3)^{2}]}{x(x - 4) + 3(x - 4)}$$

= $$\frac{5(x + 3)(x - 3)}{(x + 3)(x - 4)}$$

Here, in the numerator and denominator (x + 3) is the common factor and there is no other common factor. Now, we cancel the common factor from the numerator and denominator.

= $$\frac{5(x - 3)}{(x - 4)}$$

4. Simplify the algebraic fraction:

$$\frac{x^{4} - 13x^{2} + 36}{2x^{2} + 10x + 12}$$

Solution:

$$\frac{5x^{2} - 45}{x^{2} - x - 12}$$

Each of the numerator and denominator is polynomial, which can be factorized.

= $$\frac{x^{4} - 9x^{2} - 4x^{2} + 36}{2(x^{2} + 5x + 6)}$$

= $$\frac{x^{2}(x^{2} - 9) - 4(x^{2} - 9)}{2(x^{2} + 2x + 3x + 6)}$$

= $$\frac{(x^{2} - 4)(x^{2} - 9)}{2[x(x + 2) + 3(x + 2)]}$$

= $$\frac{(x^{2} - 4)(x^{2} - 9)}{2(x + 2)(x + 3)} [Since, a^{2} - b^{2} = (a + b)(a - b)]$$

= $$\frac{(x + 2)(x - 2)(x + 3)(x - 3)}{2(x + 2)(x + 3)}$$

Here, in the numerator and denominator (x + 2) and (x + 3) are the common factors and there is no other common factor. Now, we cancel the common factors from the numerator and denominator.

= $$\frac{(x - 2)(x - 3)(x - 3)}{2}$$



5. Reduce the algebraic fraction to its lowest term:

$$\frac{x^{2} + 5x - 2}{2x^{2} + x - 6} \div \frac{4x^{2} - 9}{6x^{2} + 7x - 3}$$

Solution:

$$\frac{x^{2} + 5x - 2}{2x^{2} + x - 6} \div \frac{4x^{2} - 9}{6x^{2} + 7x - 3}$$

Each of the numerator and denominator of each fraction are polynomial, which can be factorized.

Now by factorizing each polynomial we get;

3x2 + 5x – 2 = 3x2 –x + 6x – 2

= 3(3x – 1) + 2(3x – 1)

= (x + 2)(3x – 1)

2x2 + x – 6 = 2x2 - 3x - 4x - 6

= x(2x – 3) + 2(2x – 3)

= (x + 2)(2x - 3)

4x2 – 9 = (2x)2 - (3)2

= (2x + 3)(2x – 3)

6x2 + 7x – 3 = 6x2 – 2x + 9x – 3

= 2x(3x – 1) + 3(3x – 1)

= (2x + 3)(3x – 1)

Therefore, we have

$$\frac{(x + 2)(3x - 1)}{(x + 2)(2x - 3)} \div \frac{(2x + 3)(2x - 3)}{(2x + 3)(3x - 1)}$$

= $$\frac{(3x - 1)}{(2x - 3)} \times \frac{(2x - 3)}{(3x - 1)}$$

= $$\frac{(3x - 1)^{2}}{(2x - 3)^{2}}$$

= $$\frac{9x^{2} - 6x + 1}{4x^{2} - 12x + 9}$$

6. Reduce the algebraic fraction to its lowest form:

$$\frac{1}{x^{2} - 3x + 2} + \frac{1}{x^{2} - 5x + 6} + \frac{1}{x^{2} - 4x + 3}$$

Solution:

$$\frac{1}{x^{2} - 3x + 2} + \frac{1}{x^{2} - 5x + 6} + \frac{1}{x^{2} - 4x + 3}$$

= $$\frac{1}{x^{2} - 2x - x + 2} + \frac{1}{x^{2} - 3x - 2x + 6} + \frac{1}{x^{2} - x - 3x + 3}$$

= $$\frac{1}{x(x - 2) - 1(x - 2)} + \frac{1}{x(x - 3) - 2(x - 3)} + \frac{1}{x(x - 1) - 3(x - 1)}$$

= $$\frac{1}{(x - 2)(x - 1)} + \frac{1}{(x - 3)(x - 2)} + \frac{1}{(x - 1)(x - 3)}$$

= $$\frac{1 \times (x - 3)}{(x - 2)(x - 1)(x - 3)} + \frac{1\times (x - 1)}{(x - 3)(x - 2)(x - 1)} + \frac{1\times (x - 2)}{(x - 1)(x - 3)(x - 2)}$$

= $$\frac{(x - 3)}{(x - 2)(x - 1)(x - 3)} + \frac{(x - 1)}{(x - 3)(x - 2)(x - 1)} + \frac{(x - 2)}{(x - 1)(x - 3)(x - 2)}$$

= $$\frac{(x - 3) + (x - 1) + (x - 2)}{(x - 1)(x - 2)(x - 3)}$$

= $$\frac{(3x - 6)}{(x - 1)(x - 2)(x - 3)}$$

= $$\frac{3(x - 2)}{(x - 1)(x - 2)(x - 3)}$$

= $$\frac{3}{(x - 1)(x - 3)}$$

7. Simplify the algebraic fraction:

$$\frac{3x}{x - 2} + \frac{5x}{x^{2} - 4}$$

Solution:

$$\frac{3x}{x - 2} + \frac{5x}{x^{2} - 4}$$

= $$\frac{3x}{x - 2} + \frac{5x}{x^{2} - (2)^{2}}$$

= $$\frac{3x}{x - 2} + \frac{5x}{(x + 2)(x - 2)}$$

= $$\frac{3x \times (x + 2)}{(x - 2)(x + 2)} + \frac{5x}{(x + 2)(x - 2)}$$

= $$\frac{3x(x + 2) - 5x}{(x - 2)(x + 2)}$$

= $$\frac{3x^{2} + 6x - 5x}{(x - 2)(x + 2)}$$

= $$\frac{3x^{2} + x}{(x - 2)(x + 2)}$$

= $$\frac{x(3x + 1)}{(x - 2)(x + 2)}$$